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\title{TD du 23 janvier 2025}
\author{William Hergès\thanks{Sorbonne Université - Faculté des Sciences, Faculté des Lettres}}

\begin{document}
	\maketitle
	\newpage
	\section*{Exercice 1}
	Les produits possibles sont $AX$, $BX$, $BA$, $AB$ et $DZ$, i.e.
	$$ AX = \begin{pmatrix} 1&2\\ 2&1 \end{pmatrix}\begin{pmatrix} 1\\-1 \end{pmatrix} = \begin{pmatrix} -1\\1 \end{pmatrix} $$
	$$ BX = \begin{pmatrix} -1&1\\0&1 \end{pmatrix}\begin{pmatrix} 1\\-1 \end{pmatrix} = \begin{pmatrix} -2\\-1 \end{pmatrix} $$
	$$ BA = \begin{pmatrix} -1&1\\0&1 \end{pmatrix}\begin{pmatrix} 1&2\\2&1 \end{pmatrix}= \begin{pmatrix} 1&-1\\2&1 \end{pmatrix} $$
	$$ AB =  \begin{pmatrix} 1&2\\2&1 \end{pmatrix}\begin{pmatrix} -1&1\\0&1 \end{pmatrix} = \begin{pmatrix} -1&3\\-2&3  \end{pmatrix} $$ 
	$$ DZ = \begin{pmatrix} 1&2&3\\3&2&1\\2&1&3 \end{pmatrix} \begin{pmatrix} 0\\2\\3 \end{pmatrix} = \begin{pmatrix} 13\\7\\11  \end{pmatrix}  $$
	\section*{Exercice 2}
	\begin{enumerate}
		\item On a :
			$$ AB = \begin{pmatrix} 1&1\\0&1 \end{pmatrix} \begin{pmatrix} 0&1\\-1&0 \end{pmatrix} = \begin{pmatrix} -1&1\\-1&0 \end{pmatrix}  $$
			$$ BA =  \begin{pmatrix} 0&1\\-1&0 \end{pmatrix} \begin{pmatrix} 1&1\\0&1 \end{pmatrix}= \begin{pmatrix} 0&1\\-1&-1 \end{pmatrix}  $$
			Donc $AB \neq BA$
		\item On a :
			$$ (A + B)^2 = \begin{pmatrix} 1&2\\-1&1 \end{pmatrix} \begin{pmatrix} 1&2\\-1&1 \end{pmatrix} = \begin{pmatrix} -1&3\\-2&-1 \end{pmatrix}  $$
		\item On a :
			$$ A^2 = \begin{pmatrix} 1&1\\0&1 \end{pmatrix} \begin{pmatrix} 1&1\\0&1 \end{pmatrix} = \begin{pmatrix} 1&2\\0&1 \end{pmatrix} $$
			$$ B^2 = \begin{pmatrix} 0&1\\-1&0 \end{pmatrix} \begin{pmatrix} 0&1\\-1&0 \end{pmatrix} = \begin{pmatrix} -1&0\\0&-1 \end{pmatrix}  $$
			Après calcul, on obtient que $(A+B)^2 = A^2+AB+BA+B^2 \neq A^2+2AB+B^2$.
		\item Cherchons l'inverse de $A$ :
			$$ ad-bc = 1-0 = 1 $$
			Donc l'inverse existe, i.e.
			$$ A^{-1} = \begin{pmatrix} 1&-1\\0&1 \end{pmatrix}  $$
			Cherchons maintenant l'inverse de $B$ :
			$$ ad-bc = 0+1 = 1 $$
			Donc l'inverse existe, i.e.
			$$ B^{-1} = \begin{pmatrix} 0&-1\\1&0 \end{pmatrix}  $$
	\end{enumerate}
	\section*{Exercice 3}
	$$ A^t = \begin{pmatrix} -5&-1&3\\-4&0&4\\-3&1&5\\-2&2&6 \end{pmatrix}  $$
	puis
	$$ BA^t = \begin{pmatrix} -1&0&1&2\\ 2&1&0&-1\\ 1&0&-1&2 \end{pmatrix}A^t = \begin{pmatrix} -2&6&14\\-12&-4&4\\-8&2&10 \end{pmatrix}   $$
	\section*{Exercice 4}
	\begin{enumerate}
		\item Ce vecteur $v$ peut être représentée par un complexe $z=a+ib$. Or il existe $r,\theta\in\mathbb{R}$ tel que $z=re^{i\theta}$, d'où $r\cos\theta+ri\sin\theta$.
		\item On a 
			$$ R_{\theta}\cdot v = r(\cos\phi\cos\theta-\sin\phi\sin\theta)+ir(\sin\phi\cos\theta+\cos\theta\sin\theta) $$
			i.e.
			$$ R_{\theta}\cdot v = r\cos(\phi+\theta)+ir\sin(\phi+\theta) $$
	\end{enumerate}
	\section*{Exercice 5}
	On a :
	$$ AT = \begin{pmatrix} 2&4&8\\ 1&2&5 \end{pmatrix} $$
	$$ BT = \begin{pmatrix} 1&2&3\\-1&-2&-1 \end{pmatrix}  $$
	$$ CT = \begin{pmatrix} 0&0&2\\1&2&1 \end{pmatrix}  $$
	$$ DT = \begin{pmatrix} \frac{\sqrt{3}-1}{2}&\sqrt{3}-1&\frac{3\sqrt{3}-1}{2}\\\frac{1+\sqrt{3}}{2}&1+\sqrt{3}&\frac{3+\sqrt{3}}{2} \end{pmatrix}  $$
	\section*{Exercice 6}
	\begin{enumerate}
		\item Soit $P_n: M^n\begin{pmatrix} 0\\1 \end{pmatrix} = \begin{pmatrix} f_n\\f_{n+1} \end{pmatrix}$ avec $n\in\mathbb{N}^*$. Montrons que $P_n$ est vraie pour tout $n$.

			\fbox{Initialisation} On a $M^1\begin{pmatrix} 0\\1 \end{pmatrix} = \begin{pmatrix} 1\\1 \end{pmatrix} = \begin{pmatrix} f_{1}\\f_{2} \end{pmatrix}$. Donc $P_1$ est vraie.

			\fbox{Hérédité} Posons $n\in\mathbb{N}^*$ tel que $P_n$ soit vraie. On a :
			$$ M^n\begin{pmatrix} 0\\1 \end{pmatrix} = \begin{pmatrix} f_{n}\\f_{n+1} \end{pmatrix}  $$
			Donc,
			$$ M^{n+1}\begin{pmatrix} 0\\1 \end{pmatrix} = M\begin{pmatrix} f_n\\f_{n+1} \end{pmatrix} $$
			Or,
			$$ M\begin{pmatrix} f_n\\f_{n+1} \end{pmatrix} = \begin{pmatrix} f_{n+1}\\f_n+f_{n+1} \end{pmatrix} $$
			i.e.,
			$$ M^{n+1}\begin{pmatrix} 0\\1 \end{pmatrix} = \begin{pmatrix} f_{n+1}\\f_n+f_{n+1} \end{pmatrix} $$
			Alors,
			$$ M^{n+1}\begin{pmatrix} 0\\1 \end{pmatrix} = \begin{pmatrix} f_{n+1}\\f_{n+2} \end{pmatrix} $$
			Ainsi, $P_{n+1}$ est vraie 

			\fbox{Conclusion} $P_n$ est vraie pour tout $n\in\mathbb{N}^*$.

		\item Montrons que pour tout $n\in\mathbb{N}^*$, on a que
			$$ P_n:\quad M^n = \begin{pmatrix} f_{n-1}&f_n\\f_n&f_{n+1} \end{pmatrix}  $$
			est vraie.

			\fbox{Initialisation} On a $M^1 = \begin{pmatrix} 0&1\\1&1 \end{pmatrix} = \begin{pmatrix} f_0&f_1\\f_1&f_2 \end{pmatrix}  $. Donc $P_1$ est vraie.

			\fbox{Hérédité} Fixons $n$ dans $\mathbb{N}^*$ tel que $P_n$ soit vraie. On a :
			$$ M^n = \begin{pmatrix} f_{n-1}&f_n\\f_n&f_{n+1} \end{pmatrix} $$
			Donc,
			$$ M^nM = \begin{pmatrix} f_{n-1}&f_n\\f_n&f_{n+1} \end{pmatrix} M $$
			i.e.,
			$$ M^{n+1} = \begin{pmatrix} f_{n}&f_{n+1}\\f_{n+1}&f_n+f_{n+1} \end{pmatrix} $$
			Or,
			$$ M^{n+1} = \begin{pmatrix} f_{n}&f_{n+1}\\f_{n+1}&f_{n+2} \end{pmatrix} $$
			Ainsi, $P_{n+1}$ est vraie.

			\fbox{Conclusion} $P_n$ est vraie pour tout $n\in\mathbb{N}^*$.
	\end{enumerate}
\end{document}